Zhejiang University has 6 campuses and a lot of gates. From each gate we can collect the in/out times and the plate numbers of the cars crossing the gate. Now with all the information available, you are supposed to tell, at any specific time point, the number of cars parking on campus, and at the end of the day find the cars that have parked for the longest time period.
Input Specification:
Each input file contains one test case. Each case starts with two positive integers N (<= 10000), the number of records, and K (<= 80000) the number of queries. Then N lines follow, each gives a record in the format
plate_number hh:mm:ss status
where plate_number is a string of 7 English capital letters or 1-digit numbers; hh:mm:ss represents the time point in a day by hour:minute:second, with the earliest time being 00:00:00 and the latest 23:59:59; and status is either in or out.
Note that all times will be within a single day. Each "in" record is paired with the chronologically next record for the same car provided it is an "out" record. Any "in" records that are not paired with an "out" record are ignored, as are "out" records not paired with an "in" record. It is guaranteed that at least one car is well paired in the input, and no car is both "in" and "out" at the same moment. Times are recorded using a 24-hour clock.
Then K lines of queries follow, each gives a time point in the format hh:mm:ss. Note: the queries are given in ascending order of the times.
Output Specification:
For each query, output in a line the total number of cars parking on campus. The last line of output is supposed to give the plate number of the car that has parked for the longest time period, and the corresponding time length. If such a car is not unique, then output all of their plate numbers in a line in alphabetical order, separated by a space.
Sample Input: 16 7 JH007BD 18:00:01 in ZD00001 11:30:08 out DB8888A 13:00:00 out ZA3Q625 23:59:50 out ZA133CH 10:23:00 in ZD00001 04:09:59 in JH007BD 05:09:59 in ZA3Q625 11:42:01 out JH007BD 05:10:33 in ZA3Q625 06:30:50 in JH007BD 12:23:42 out ZA3Q625 23:55:00 in JH007BD 12:24:23 out ZA133CH 17:11:22 out JH007BD 18:07:01 out DB8888A 06:30:50 in 05:10:00 06:30:50 11:00:00 12:23:42 14:00:00 18:00:00 23:59:00 Sample Output: 1 4 5 2 1 0 1 JH007BD ZD00001 07:20:09题意：一共有n个记录，记录车的进出时间，车牌号，要忽略一些无效记录（每辆车的进出记录应该一对对出现）。m次询问，每次给出一个时间，问这个时间又几辆车在车库。最后输出在车库待的时间最久的车，并输出时长
解题思路：先将所有记录排一下序，车牌号相同的按时间点来排，不同的按字典序排，遍历一遍记录，计算出每个时间又几辆车，并用map记录每辆车停留的时间
#include <iostream> #include <cstdio> #include <cstring> #include <string> #include <algorithm> #include <vector> #include <queue> #include <stack> #include <map> #include <set> #include <cmath> using namespace std; #define LL long long const int INF = 0x3f3f3f3f; struct node { char s[20],ss[5]; int h,m,c,sum; friend bool operator <(node a,node b) { if(!strcmp(a.s,b.s)) return a.sum<b.sum; else return strcmp(a.s,b.s)<0; } } a[10008]; int n,m,sum[100000]; map<string,int>mp; vector<string>g; int main() { while(~scanf("%d%d",&n,&m)) { mp.clear(); g.clear(); memset(sum,0,sizeof sum); for(int i=1; i<=n; i++) { scanf("%s%d:%d:%d%s",a[i].s,&a[i].h,&a[i].m,&a[i].c,a[i].ss); a[i].sum=a[i].h*3600+a[i].m*60+a[i].c; } sort(a+1,a+1+n); int ma=0; for(int i=2;i<=n;i++) { if(!strcmp(a[i].ss,"out")&&!strcmp(a[i-1].ss,"in")&&!strcmp(a[i].s,a[i-1].s)) { sum[a[i].sum]--;sum[a[i-1].sum]++; ma=max(ma,mp[a[i].s]+=a[i].sum-a[i-1].sum); } } for(int i=1;i<=24*3600;i++) sum[i]+=sum[i-1]; while(m--) { int h,m,s; scanf("%d:%d:%d",&h,&m,&s); printf("%d\n",sum[h*3600+60*m+s]); } for(map<string,int>::iterator it=mp.begin();it!=mp.end();it++) if(it->second==ma) g.push_back(it->first); sort(g.begin(),g.end()); int Size=g.size(); for(int i=0;i<Size;i++) cout<<g[i]<<" "; printf("d:d:d\n",ma/3600,(ma600)/60,ma`); } return 0; }
